Two taps can fill a cistern in 20 minutes and 30 minutes. The first

I backsolved here. Finding equation was taking too long.

1. Start with (C), where Tap 1 is open by itself for 9 minutes.

Tap 1 rate = \(\frac{1}{20}\)

W = rt, so \(\frac{1}{20}\) * 9 = \(\frac{9}{20}\) work is completed --->

\(\frac{11}{20}\) work remains.

9 minutes have passed.

2. Then Tap 1 and Tap 2 work together. Add rates.

\(\frac{1}{20}\) + \(\frac{1}{30}\) = \(\frac{50}{600}\) = \(\frac{1}{12}\)

W/r = t: \(\frac{11}{20}\) divided by \(\frac{1}{12}\) = \(\frac{132}{60}\) = 6.6 minutes

9 minutes elapsed + 6.6 more = 15.6 minutes. x = 9 is just a little too large. Try (B).

3. Answer (B), Tap 1 alone for 7.5 = \(\frac{15}{2}\) minutes

Work done by Tap 1: \(\frac{1}{20}\) * \(\frac{15}{2}\) = \(\frac{3}{8}\)

\(\frac{5}{8}\) of work remains. 7.5 minutes have elapsed.

4. Taps 1 and 2 work together. From above, combined rate is \(\frac{1}{12}\).

W/r = t: \(\frac{5}{8}\) divided by \(\frac{1}{12}\) =

\(\frac{15}{2}\) = 7.5 minutes.

7.5 minutes elapsed + 7.5 minutes more = 15 minutes. Answer B.

Assume volume of the tank to be 600 units.
First tap takes 20 minutes(Rate : 30 units/minute)
Second tap takes 30 minutes(Rate : 20 units/minute)

Given data : The first tap was opened initially for x minutes
Total time taken to fill the tank was 15 minutes
Since we know that the taps combined fill (20+30 units/minute = 50 units/minute)

From question stem, we can frame equation
30x + 50(15-x) = 600

Solving for x,
30x + 750 - 50x = 600
750 - 600 = 50x - 30x
150 = 20x
x = 7.5 minutes(Option B)

Responding to a pm:

Another way to look at it:

Combined rate of both taps = 1/20 + 1/30 = 5/60 = 1/12

If both worked together, it would have taken 12 mins. 3 mins less will be taken if the second tap also works for the whole time. So in 3 mins, both taps do 1/4 of the work. That is what the second tap would have done extra (if it were open for the whole time)

Time in which second tap did 1/4 of the work = (1/4)/(1/30) = 7.5 mins

So the second tap was on for 7.5 mins and the first tap was on alone for 15-7.5 = 7.5 mins = x

Answer (B)

\(\frac{x}{20} + 15-x(\frac{1}{20}+\frac{1}{30}) = 1\)

\(\frac{x}{20} + \frac{15-x}{20} + \frac{15-x}{30} = 1\)

\(\frac{15}{20} + \frac{15-x}{30} = 1\)

\(450 + 300 - 20x = 600\)

\(20x = 150\)

\(x = 7.5\). Ans - B.

We are given that the rate of the first cistern is 1/20 and the rate of the second cistern is 1/30. The first tap worked for 15 minutes and the second tap worked for (15 - x) minutes. Thus:

(1/20)(15) + (1/30)(15 - x) = 1

3/4 + (15 - x)/30 = 1

Multiplying the entire equation by 60, we have:

45 + 30 - 2x = 60

2x = 15

x = 7.5

Answer: B

Given:
1. Two taps can fill a cistern in 20 minutes and 30 minutes.
2. The first tap was opened initially for x minutes after which the second tap was opened.

Asked: If it took a total of 15 minutes for the tank to be filled, what is the value of x?

15/20 + (15-x)/30 = 1
15-x = 5*30/20 = 7.5
x = 7.5

IMO B

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